Question

You drop a rock from the top of a building of height h. Your co-experimenter throws a rock from the same spot with a vertically downward speed vo, a time t after you released your rock. The two rocks hit the ground at the same time. Find the expression for the time t, in terms of vo, g, and h.

Answer 1

`t=√(2h/g)-(1/g)*(\sqrt{v_(o)^2+2gh}-v_(o))`

Step-by-step explanation:

First person:

`y(t)=y_(o)-v_(o)t-1/2*g*t^(2)`

`v_(o)=0` the rock is dropped

`y_(o)=h`

`y(t)=h-1/2*g*t^(2)`

after t1 seconds it hit the ground, y(t)=0

`0=h-1/2*g*t_(1)^(2)`

`t_(1)=√(2h/g)`

Second person:

`y(t)=y_(o)-v_(o)t-1/2*g*t^(2)`

`v_(o)` the rock has a initial downward speed

`y_(o)=h`

`y(t)=h-v_(o)t-1/2*g*t^(2)`

after t2 seconds it hit the ground, y(t)=0

`0=h-v_(o)t_(2)-1/2*g*t_(2)^(2)`

`g*t_(2)^(2)+2v_(o)t_(2)-2h=0`

`t_(2)=(1/2g)*(-2v_(o)+\sqrt{4v_(o)^2+8gh})`

the time t when the second person throws the rock after the first person release the rock is:

t=t1-t2

`t=√(2h/g)-(1/g)*(\sqrt{v_(o)^2+2gh}-v_(o))`